Does toothpaste have matter

9) A balloon is charged with 3.4 μC (microcoulombs) of charge. A second balloon 23 cm away is charged with -5.1 μC of charge. Th

lina2011 [118]

Answer:

9. The force is a force of attraction and it is 2.95N

10. The magnitude of acceleration 35.12m/s^2 and the direction of this acceleration is away from the other balloon.

Explanation:

Parameters given:

Q1 = 3.4 * 10^-6C

Q2 = - 5.1 * 10^-6C

Distance between the two balloons = 23cm = 0.23m

9. Force acting between the two balloons is a force of attraction because they are unlike charges. Hence, the force between them is:

F = kQ1Q2/r^2

F = (9 *10^9 * 3.4 * 10^-6 * -5.1 * 10^-6)/(2.3 * 10^-1)^2

F = (1.56 * 10^-1)/(5.29 * 10^-2)

F = - 2.95N

10. Assuming that Balloon A has a mass, m, of 0.084kg, then:

F = ma

Where a = acceleration

a = F/m

a = -2.95/0.084

a = - 35.12m/s^2

The acceleration has a magnitude of 35.12m/s^2 and its direction is away from balloon B.

The negative sign shows that the balloon A is slowing down as it moves towards balloon B. Hence, it's velocity is reducing slowly.

4 0

4 years ago

A 2.90 kg mass is pushed against a horizontal spring of force constant 23.0 n/cm on a frictionless air table. the spring is atta

cricket20 [7]

A boiling pot of water (the water travels in a current throughout the pot), a hot air balloon (hot air rises, making the balloon rise) , and cup of a steaming, hot liquid (hot air rises, creating steam) are all situations where convection occurs.
Read more on Brainly.com - brainly.com/question/1581851#readmore

7 0

3 years ago

A waitperson carrying a tray with a platter on it tips the tray at an angle of 12 degrees below the horizontal. If the gravitati

Tasya [4]

Answer:1.04 N

Explanation:

Given

Gravitational Force on the Platter is $5 N$

Tray makes an angle of $\theta =12^{\circ}$

This gravitational Force has components along and Perpendicular to Platter

Perpendicular Force $W_p=W\cos \theta$

$W_p=5\times \cos 12=4.89 N$

Along the Tray

$W_{along}=W\sin \theta$

$W_{along}=5\times \sin 12=1.04 N$

Thus 1.04 N is the magnitude of force that will cause Platter to slide down

7 0

3 years ago

The acceleration due to gravity on the surface of Mars is about one-third the acceleration due to gravity on Earthâ€™s surface.

aksik [14]

Answer:

one-third of its weight on Earth's surface

Explanation:

Weight of an object is = W = m*g

Gravity on Earth = g₁ = 9.8 m/s

Gravity on Mars = g₂ = $\frac{1}{3}$ g₁

Weight of probe on earth = w₁ = m * g₁

Weight of probe on Mars = w₂ = m * g₂ -------- ( 1 )

As g₂ = g₁/3 --------- ( 2 )

Put equation (2) in equation (1)

so

Weight of probe on Mars = w₂ = m * g₁ /3

Weight of probe on Mars = $\frac{1}{3}$ m * g₁ = $\frac{1}{3}$ w₁

⇒Weight of probe on Mars = $\frac{1}{3}$ Weight of probe on earth

6 0

4 years ago

A stone is thrown horizontally from 2.4m above the ground at 35m/s. The wall is 14m away and 1m high.At what height the stone wi

KIM [24]

The stone reaches the wall at a height of 1.62 m.

The stone lands at a point 24.5 m from the point of projection.

The stone is projected horizontally with a velocity u at a height h from the ground. The wall is located at a distance x from the point of projection. The stone takes a time t to reach the wall and in the same time the stone falls a vertical distance y.

The horizontal distance x is traveled with a constant velocity u.

$x=ut$

Calculate the time taken t.

$t=\frac{x}{u} \\ =\frac{14m}{35 m/s} \\ =0.40s$

The stone's initial vertical velocity is zero. It falls through a distance y in the time t under the action of acceleration due to gravity g.

$y=\frac{1}{2} gt^2\\ \frac{1}{2} (9.81m/s^2)(0.40s)^2\\ =0.784m$

The height h₁ of the stone above the ground when it reaches the wall is given by,

$h_1=h-y\\ =(2.4m)-(0.784m)\\ =1.616m=1.62m$

When the stone reaches the wall, its height from the ground is 1.62m.

The stone thus crosses over the wall, since the height of the wall is 1 m. It reaches the ground at a distance R from the point of projection. If the time taken by the stone to reach the ground is t₁, then,

$h=\frac{1}{2} gt_1^2$

Calculate the time taken by the stone to reach the ground.

$t_1=\sqrt{\frac{2h}{g} } \\=\sqrt{\frac{2(2.4m)}{9.81m/s^2} } \\ =0.699 s$

The horizontal distance traveled by the stone is given by,

$R=ut_1 \\ =(35m/s)(0.699s)\\ =24.5m$

The stone lands at point 24.5 m from the point of projection and 10.5 m from the wall.

3 0

3 years ago